The probability density function of X is given by fx a bx2

The probability density function of X is given by f(x) = {a + bx^2, 0

Solution

We know E(x) = x.f(x) dx

Given that E(x) = 3/5

E(x) = x*f(x) dx     =3/5

E(x) = x*(a+bx²) dx = 3/5

E(x) = (ax) dx + (bx²) dx = 3/5

E(x) = ax dx + bx² dx = 3/5

E(x) = a[x²/2]¹₀ + b[x³/3]₀¹ = 3/5

E(x) = (a/2) + (b/4) =3/5

10a+5b = 3/5 1

We know    f(x) dx = 1

(a+bx²) dx = 1

a dx+ (bx²) dx   = 1

a dx+ b x² dx   = 1

a+ (b/3) = 1

3a+b = 3 2

Solve 1 and 2 equation

a = 3/5    b = 6/5